Problem:
Let P(x)=x2−3x−7, and let Q(x) and R(x) be two quadratic polynomials also with the coefficient of x2 equal to 1. David computes each of the three sums P+Q,P+R, and Q+R and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If Q(0)=2, then R(0)=nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
We are given that P+Q, P+R, and Q+R all have a common root, respectively.
Notice that since each of these quadratics has no leading coefficient, that adding pairs would have a leading coefficient of 2.
Suppose the common root of P+Q and P+R is a, P+Q and Q+R is is b, and P+R and Q+R is c.
Then the polynomials can be written as following:
P+Q=2(x−a)(x−b)
Q+R=2(x−b)(x−c)
P+R=2(x−a)(x−c)
Then we'd have that, adding these all up that
21(P+Q+R)=(x−a)(x−b)+(x−a)(x−c)+(x−b)(x−c)
Now, notice that we can write P(X), not directly but as a combination of expressions we have, in particularly substiting
P(x)=(P(x)+Q(x)+R(x))−(Q(x)+R(x))
P(x)=(x−a)(x−b)+(x−a)(x−c)+(x−b)(x−c)−2(x−b)(x−c)
=(x−a)(x−b)+(x−a)(x−c)+(x−b)(x−c)−2(x−b)(x−c)
=x2−2ax+ab+ac−bc
Since P(x)=x2−3x+7, we can solve −2a=−3→a=23.
From here, we have 3 equations, doing the same thing we did for P(x) for Q and R.
We'd have that
With P(0), we'd have that ab+ac−bc=−7
Looking at Q(0) in particular, we'd have that Q(0)=−2=ab+bc−ac
This yields
a=23,b=−35,c=−1927
Plugging this in for R(0) yields
R(0)=1952→52+19=71
The problems on this page are the property of the MAA's American Mathematics Competitions